package com.gxc.linkedList;

/**
 * 21. 合并两个有序链表
 * 将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
 */
public class MergeTwoLists {

    public static void main(String[] args) {

    }

    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null) return list2;
        if (list2 == null) return list1;
        ListNode head = list1.val<list2.val ? list1 : list2;
        ListNode node = head;
        ListNode next = null;
        while (list1 != null && list2 != null) {
            if (list1.val<list2.val) {
                next = list1;
                list1 = list1.next;
            } else {
                next = list2;
                list2 = list2.next;
            }
            if (next == head) continue;
            node.next = next;
            node = next;
        }
        if (list1 != null) {
            node.next = list1;
        }
        if (list2 != null) {
            node.next = list2;
        }
        return head;
    }

    public class ListNode {
          int val;
          ListNode next;
          ListNode() {}
         ListNode(int val) { this.val = val; }
          ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }

    //官方解
    class Solution {
        public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            //用虚拟头结点 虚拟头.next 就是 head
            ListNode prehead = new ListNode(-1);

            ListNode prev = prehead;
            while (l1 != null && l2 != null) {
                if (l1.val <= l2.val) {
                    prev.next = l1;
                    l1 = l1.next;
                } else {
                    prev.next = l2;
                    l2 = l2.next;
                }
                prev = prev.next;
            }

            // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
            prev.next = l1 == null ? l2 : l1;

            return prehead.next;
        }
    }


    public ListNode mergeTwoLists3(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        } else if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }




}
